Question

Integrate sin⁶ 2x dx limit [0,π/4]​???​

Answer 1

$\large\underline{\sf{Solution-}}$

Let assume that

$\rm \longrightarrow \: I = \displaystyle \int_{0}^{ \dfrac{\pi}{4} } \rm \: {sin}^{6} 2x \: dx$

Now we use here, substitution method

$\red{\rm :\longmapsto\:Put \: 2x = y} \\ \red{\rm :\longmapsto\:2dx = dy} \\ \red{\rm :\longmapsto\:dx = \dfrac{dy}{2}}$

When we use substitution method, we have to change the limits also.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf x & \bf y = 2x \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf 0 & \sf 0 \\ \\ \sf \dfrac{\pi}{4} & \sf \dfrac{\pi}{2} \end{array}} \\ \end{gathered}$

So, given integral reduced to

$\rm \longrightarrow \: I = \displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {sin}^{6} y \: \dfrac{dy}{2}$

$\rm \longrightarrow \: I = \dfrac{1}{2} \displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {sin}^{6} y \: dy$

We know,

Walli's Formula, if n is even natural number,

$\rm \longrightarrow\:\displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {sin}^{n} x \: dx = \frac{(n - 1)(n - 3) - - - 1}{n(n - 2)(n - 4) - - 2} \times \dfrac{\pi}{2}$

So, apply this formula, we get

$\rm \longrightarrow \: I = \dfrac{1}{2} \times \dfrac{5.3.1}{6.4.2} \dfrac{\pi}{2}$

$\rm \longrightarrow \: I = \dfrac{5\pi}{64}$

Hence,

$\rm \longrightarrow \:\displaystyle \int_{0}^{ \dfrac{\pi}{4} } \rm \: {sin}^{6} 2x \: dx \: = \: \dfrac{5\pi}{64}$

Additional Information :-

1. If n is even natural number, then

$\rm \longrightarrow\:\displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {sin}^{n} x \: dx = \frac{(n - 1)(n - 3) - - - 1}{n(n - 2)(n - 4) - - 2} \times \dfrac{\pi}{2}$

and

$\rm \longrightarrow\:\displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {cos}^{n} x \: dx = \frac{(n - 1)(n - 3) - - - 1}{n(n - 2)(n - 4) - - 2} \times \dfrac{\pi}{2}$

2. If n is odd natural number, then

$\rm \longrightarrow\:\displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {cos}^{n} x \: dx = \frac{(n - 1)(n - 3) - - - 2}{n(n - 2)(n - 4) - - 1}$

and

$\rm \longrightarrow\:\displaystyle \int_{0}^{ \dfrac{\pi}{2} } \rm \: {sin}^{n} x \: dx = \frac{(n - 1)(n - 3) - - - 2}{n(n - 2)(n - 4) - - 1}$